/* Medium
Given a root node reference of a BST and a key,
delete the node with the given key in the BST.
Return the root node reference (possibly updated) of the BST.

Basically, the deletion can be divided into two stages:

Search for a node to remove.
If the node is found, delete the node.

Example 1:
Input: root = [5,3,6,2,4,null,7], key = 3
Output: [5,4,6,2,null,null,7]
Explanation: Given key to delete is 3. So we find the node with value 3 and delete it.
One valid answer is [5,4,6,2,null,null,7], shown in the above BST.
Please notice that another valid answer is [5,2,6,null,4,null,7] and it's also accepted.

Example 2:
Input: root = [5,3,6,2,4,null,7], key = 0
Output: [5,3,6,2,4,null,7]
Explanation: The tree does not contain a node with value = 0.
Example 3:

Input: root = [], key = 0
Output: []

Constraints:
The number of nodes in the tree is in the range [0, 10^4].
-10^5 <= Node.val <= 10^5
Each node has a unique value.
root is a valid binary search tree.
-10^5 <= key <= 10^5 */

struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode() : val(0), left(nullptr), right(nullptr) {}
    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
    TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
};

class Solution {
public:
    TreeNode* deleteNode(TreeNode* root, int key) {
        if (!root)
            return nullptr;

        if (root->val == key) {
            if (!root->left || !root->right) {
                return root->left ? root->left : root->right;
            } else {
                auto minNode = getMin(root->right);
                root->val = minNode->val;
                root->right = deleteNode(root->right, minNode->val);
            }
        } else if (root->val < key) {
            root->right = deleteNode(root->right, key);
        } else {
            root->left = deleteNode(root->left, key);
        }

        return root;
    }

private:
    TreeNode* getMin(TreeNode* node) {
        while (node->left) {
            node = node->left;
        }
        return node;
    }
};